Definition. A sequence is a function . It is usual to represent the sequence f as where .

Definition. A sequence is said to converge to a if for each there is an such that

for all .

It can be shown that this number is unique and we write .

Examples.

(a) The sequence converges to 0. For , let . This gives

for all .

(b) The sequence converges to 2. The nth term of this sequence is . So . But, for all . Thus, for a given , the choice of given in (a) above will do.

(c) The sequence defined by converges to 1.

Let so that for , . Now, , thus, for , we have . For any , we can find in N such that . Thus, for any , we have . This is the same as writing

for or equivalently,

(d) Let . The sequence converges to o. Note that

for .

For , choose such that . Now, let so that for all .

(e) For , the sequence converges to 0. (Exercise!)

In all the above examples, we somehow guessed in advance what a sequence converges to. But, suppose we are not able to do that and one asks whether it is possible to decide if the sequence converges to some real number. This can be helped by the following analogy: Suppose there are many people coming to Delhi to attend a conference. They might be taking different routes. But, as soon as they come closer and closer to Delhi the distance between the participants is getting smaller.

This can be paraphrased in mathematical language as:

Theorem. If is a convergent sequence, then for every , we can find an such that

for all .

Proof. Suppose converges to a. Then, for every we can find an such that

for all .

So, for , we have

. QED.

The proof is rather simple. This very useful idea was conceived by Cauchy and the above theorem is called Cauchy criterion for convergence. What we have proved above tells us that the criterion is a necessary condition for convergence. Is it also sufficient? That is, given a sequence which satisfies the Cauchy criterion, can we assert that there is a real number to which it converges? The answer is yes!

We call a sequence monotonically non-decreasing if for all n.

Theorem. A monotonically non-decreasing sequence which is bounded above converges.

Proof. Suppose is monotonic and non-decreasing. We have . Since the sequence is bounded above, has a least upper bound. Let it be a. By definition, for all n, but for there is at least one such that . Therefore, for , . This gives us for all . QED.

We can similarly prove that: A monotonically non-increasing sequence which is bounded below is convergent.

Suppose we did not have the condition of boundedness below or above for a monotonically non-increasing or non-decreasing sequence respectively, then what would happen? If a sequence is monotonically non-decreasing and is not bounded above, then given any real number there exists at least one such that , and hence for all . In such a case, we say that diverges to . We write . More generally, (that is, even when the sequence is not monotone), the same criterion above allows us to say that diverges to and we write . We can similarly define divergence to .

We make a digression here: In case a set is bounded above, then we have the concept of least upper bound. For any set A or real numbers, we define supremum of A as

, if A is bounded above and is equal to if A is not bounded above.

Similarly, we define infimum of a set A as

, if A is bounded below, and is equal to , if A is not bounded below.

This would help us to look for other criteria for convergence. For any bounded sequence (a_{n})_{n=1}^{\infty} of real numbers, let

.

It is clear that is now a non-increasing sequence. So, it converges. We set

,

Similarly, if we write

,

then is a monotonically non-decreasing sequence. So we write

,

We may not know a sequence to be convergent or divergent, yet we can find its limit superior and limit inferior.

We have in fact the following result:

Theorem.

For a sequence of real numbers, .

Further, if , and are finite, then if and only if the sequence is convergent.

Proof.

It is easy to see that . Now, suppose l, L are finite.If are finite. If , then for all there exists such that

for all and for all .

Thus, with we have

for all .

This proves that equality holds only when it is convergent. Conversely, suppose converges to a. For every , we have such that for all . Therefore, and

Let . If , then and hence . On the other hand, if , then . Thus, A is non-empty. Next, observe that if , then v is an upper bound of A. In particular, is an upper bound of A.

By the least upper bound property, A admits a least upper bound. Let us denote it by y. We will rule out the possibilities and implying that .

If , let and . It can be checked that so that . But, , contradicting the fact that y is the least upper bound of A. (we have used the inequalities 7 and 8 in the previous blog).

On the other hand, if , let and . Again, it can be verified that and hence w is an upper bound of A. But, , contradicting the fact that y is the least upper bound of A. Hence, . QED.

In particular, we see that there is an element such that and hence also which means that the equation has two solutions. The positive one of those two solutions is . In fact, the above theorem has guaranteed its extraction of the square root, cube root, even nth root of any positive number. You could ask at this stage, if this guarantees us extraction of square root of a negative number. The answer is clearly no. Indeed, we have

for .

Remark.

We can further extend to include numbers whose squares are negative, which you know leads to the idea of complex numbers.

We have shown that Q is a subset of . One can also show that between any two distinct real numbers there is a rational number. This naturally leads to the decimal representation of real number: Given any real number x and any , we can get a unique and unique such that and

You are invited to try to prove this familiar decimal representation.

If we have a terminating decimal representation of a real number, then surely it is rational.But, we know that rationals like 1/3, 1/7, 1/11, do not have a terminating decimal expansion.

It is clear that the decimal representation of cannot terminate as it is not rational. There are many elements of which are not in Q. Any element of which is not in Q is called an irrational number, and irrational numbers cannot have terminating decimal representation.

Our scheme of real numbers has so far been extended to rational numbers so that we could not only add and subtract without any hindrances, but could also multiply and divide without any restriction, save division by zero. But, when it comes to extraction of square roots, it seems “incomplete”. For example, there is no rational number whose square gives 2. This question crops up when we try to relate numbers to geometry.

We have seen before that we can always represent a rational number on a line. Now suppose we have a square with sides of unit length. By the theorem of Pythagoras, the square of the length of diagonal is equal to . We can also see that the length of the diagonal can be geometrically represented on the line. Now, we ask here, does there exist a rational number whose square is 2, or equivalently, does the equation have a solution in rational numbers? This amounts to asking if we can find two non-zero integers p and q such that

. Equation I.

We can demonstrate that there are no such integers. Suppose that there were two such integers p and q satisfying equation I. We may assume without loss of generality, that p and q have no common factors. If there were any, we could cancel them from both the sides of equation I till there were no common factors. Now as there are no common factors between p and q, for equation I to hold, 2 must be a factor of , and hence, of p. So we may write

Equation 2.

Substituting this in Equation 2, we get . Equation 3.

As before, we conclude that for equation 3 to hold, 2 must divide q giving for some integer n. This contradicts our assumption that p and q have no common factors. This only proves that equation I has no solution in integers (except ). Thus, the length of the diagonal of the unit square, though it has a point representing it on the line, does not correspond to a rational number. This shows that Q is not large enough to accommodate a number such as the length of the diagonal of a unit square. So we must extend Q. We could simply include the new number which we write as in the scheme along with Q. But, then this ad hoc extension may not stand up to all our demands to include newer and newer numbers which arise out of algebraic equations, e.g.,, etc. Even if we somehow include these numbers in our scheme, we must know how to perform arithmetic operations like addition, subtraction, multiplication and division in it. Also, how does one decide which of a given pair of new numbers is smaller? There are many ways in extending Q to accommodate all these new numbers. We illustrate one of the simplest ways of doing this.

Let . It is clear that E has an upper bound, for example, 2. Next, we note that if is an upper bound of E, then there exists with (we can always choose an n because of the Archimedean property of Q). Then, one has , showing that and thus no upper bound of E can be in E. Let be one such upper bound of E, then so is since . On the other hand, since . Therefore, the E has no least upper bound in Q.

This tells us what to do. We now extend the field of rationals to a larger field containing it which has all properties of Q along with an additional property, called the least upper bound property and abbreviated as lubproperty, namely, every set that is bounded above has a least upper bound.

More precisely, we postulate that there is a field , containing Q and satisfying all the properties of Q listed above and an additional one:

The least upper bound property (lub): If is bounded above, then A admits a least upper bound, that is, (1) there is a such that for every and (2) for every , there is a such that .

Remark:

As we constructed Q out of Z, there are ways of constructing out of Q. However, an explicit construction of from Q is beyond the scope of the present blog. Nevertheless, we are going to use the listed properties of in all subsequent discussions and deduce many interesting consequences.

Since has all the properties of Q including the order property, we write (as for Q),

and this has the same properties as . Now, we use the same notation like for elements of , exactly as we did for the elements of Q. As before, we write or according as or .

Clearly, 1, belong to and represent the natural numbers. So, one sets up a one to one correspondence between and to conclude that . Writing as the additive inverse of , of course , we see that . Since is a field, for , , we must have

where is the multiplicative inverse of n. We agree to write . This way we have . We have thus effectively extended Q to .

We have seen earlier that every element of Z can be represented on a straight line with a certain point representing 0. Positive integers are represented by points to the right of this marked point at equal lengths, and negative integers are similarlyrepresented by points to the left of this.

We can use the same straight line to represent the newly constructed rational numbers as follows.

We know that we can divide a line segment, using a straight edge and compass, into q equal parts. So, the line segment between 0 to 1 can be divided into q equal parts for every positive integer q.

The points of division will now represent the numbers . It is now clear that any rational number of the form can be represented on the straight line. We have seen that between any two distinct rational numbers r and , no matter how close they are, we can always find another rational number between them, example, . Geometrically, this means that between any two distinct points on the line representing rational numbers, there is a point between them representing a rational number.

One can ask: does every point on the line correspond to a rational number? Note that we also have a point on the line representing length of the diagonal of a square of side 1. But, we shall show in the next blog that the length of the diagonal of unit square does not correspond to a rational number.

To prepare the ground for such discussions, let us start with a few definitions:

A subset is said to be bounded below if there exists a rational number such that

for every

and is called a lower bound of A. Similarly, is said to be bounded aboveif there is a such that , and is called an upper bound of A. A set which is bounded above and also bounded below is called a bounded set. Observe that if a set has a lower bound then it has many more like , . Similarly, a set that is bounded above has many upper bounds. If there is a least element of all these upper bounds, then we call it the least upper bound of the set. But, it is not true that every set of rational numbers which is bounded above has a least upper bound. We can say similar things about sets which are bounded below. Some sets may have least upper bound. For example, the set of negative rational numbers is certainly bounded above and it is easy to see that 0 is the least upper bound.